Problem: Graph this system of equations and solve. $-x-5y = 20$ $-9x-5y = -20$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Answer: Convert the first equation, $-x-5y = 20$ , to slope-intercept form. $y = -\dfrac{1}{5} x - 4$ The y-intercept for the first equation is $-4$ , so the first line must pass through the point $(0, -4)$ The slope for the first equation is $-\dfrac{1}{5}$ . Remember that the slope tells you rise over run. So in this case for every $1$ position you move down (because it's negative) You must also move $5$ positions to the right. $5$ positions to the right. Graph the blue line so it passes through $(0, -4)$ and $(5, -5)$ Convert the second equation, $-9x-5y = -20$ , to slope-intercept form. $y = -\dfrac{9}{5} x + 4$ The y-intercept for the second equation is $4$ , so the second line must pass through the point $(0, 4)$ The slope for the second equation is $-\dfrac{9}{5}$ . Remember that the slope tells you rise over run. So in this case for every $9$ positions you move down (because it's negative) You must also move $5$ positions to the right. $5$ positions to the right. $9$ positions down from $(0, 4)$ is $(5, -5)$ Graph the green line so it passes through $(0, 4)$ and $(5, -5)$ The solution is the point where the two lines intersect. The lines intersect at $(5, -5)$.